Every $p$-subgroup is contained in one $p$-Sylow subgroup?

I am learning Sylow's theorems in my algebra course and I was reading questions posted before. One is the following: If $H$ is a normal subgroup of a finite group $G$ and $|H|=p^k$ for some prime $p$. show that $H$ is contained in every sylow $p$ subgroup of $G$

$H$ is subgroup of a power of $p$, so it is contained in a maximal $p$-subgroup. Indeed, the maximal $p$-subgroup, call it $A$, is a $p$-Sylow subgroup. If $B$ is another $p$-Sylow subgroup, then by Sylow's 2nd theorem, they are conjugate of each other. Since $H$ is normal, the conjugate of $H \subset A$ is $H$, which is contained in $B$. Thus, $H$ is contained in every $p$-Sylow subgroup of $G$.

It seems fine but I don't see why the bolded words stand. It may be trivial to you but I would be thankful if you can show it in some details. Thanks in advance.

• have a look at this Nov 7 '14 at 16:27
• Yes I have included this link in my post but I still don't understand this fine detail after reading it. Nov 7 '14 at 16:29
• have a look at the second proof it says $HK$ is a subgroup where $H$ is the given subgroup and $K$ is taken to be the Sylow P Subgroup now order of $HK$ is more than $P^m$ if $H \cap K \neq H$ where m is the max power of p dividing $|G|$ Nov 7 '14 at 16:31

A subset of a set is said to be "maximal" with respect to a given property if it is not contained in a different, generally larger, subset having the property. There a maximal $p$ subgroup is a $p$ subgroup that is not contained in another $p$ subgroup.
We could use Zorn's lemma for this, but note that the group is finite. If the $p$-group is not contained in another $p$-group, then it is maximal. Otherwise it is contained in a larger $p$-group, and you can use descending induction to show that since the larger $p$-group is contained in a maximal $p$-group, so is the smaller one. By definition a maximal $p$ subgroup is a Sylow subgroup, so every $p$-subgroup is contained in a Sylow subgroup.